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\title{Applied Stochastic Processes - Lecture 08}
\subtitle{(8.1 - 8.2) Brownian Motions -  the Reflection Principle}
%(3.1-3.3) 
%\institute{上海立信会计金融学院}
\author{MAP SK}
\date{May 25, 2021}

\maketitle

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%\begin{enumerate}
%\item 第一讲：March 9 (3.1-3.3) Markov Chains - Concepts and Examples
%\item 第二讲：March 16 (3.4-3.5) Markov Chains - First Step Analysis, More Examples
%\item 第三讲：March 23 (4.1-4.4) Markov Chains - Long Run Behaviors
%\item 第四讲：April 13 (5.1-5.2) Poisson Processes - Concept and Examples 
%\item 第五讲：April 20 (5.3-5.4) Poisson Processes - Associated Distributions  
%\item 第六讲：May 11 (6.1-6.1) Markov Chains - Continuous Time, Pure Birth Processes
%\item 第七讲：May 18 (7.1-7.2) Renewal Processes - Renewal Function, Block Replacement
%\item 第八讲：May 25 (8.1-8.2) Brownian Motions -  the Reflection Principle 
%\end{enumerate}

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\begin{frame}{Contents }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{itemize}
\item  A little history
\item  The Brownian motion stochastic process
\item  The diffusion equation
\item  The definition of Brownian motion
\item  The maximum variable and the reflection principle
\end{itemize}

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%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{8.1. A Little History }

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\begin{itemize}

\item  The Brownian motion stochastic process arose early in this century as an attempt to explain the ceaseless irregular motions of tiny particles suspended in a fluid, such as dust motes floating in air. 

\item  Today, the Brownian motion process and its many generalizations and extensions occur in numerous and diverse areas of pure and applied science such as economics, communication theory, biology, management science, and mathematical statistics.

\item  The story begins in the summer of 1827, when the English botanist Robert Brown observed that microscopic pollen grains suspended in a drop of water moved constantly in haphazard zigzag trajectories. 

\item  Following the reporting of his findings, other scientists verified the strange phenomenon. 

\end{itemize}

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\begin{itemize}


\item  Similar Brownian motion was apparent whenever very small particles were suspended in a fluid medium, e.g., smoke particles in air. 

\item  Over time, it was established that finer particles move more rapidly, that the motion is stimulated by heat, and that the movement becomes more active with a decrease in fluid viscosity.

\item  A satisfactory explanation had to wait until the next century, when in 1905, Einstein would assert that the Brownian motion originates in the continual bombardment of the pollen grains by the molecules of the surrounding water, with successive molecular impacts coming from different directions and contributing different impulses to the particles. 

\item  Einstein argued that as a result of the continual collisions, the particles themselves had the same average kinetic energy as the molecules. 

\end{itemize}

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%\begin{itemize}
%
%
%\item  Belief in molecules and atoms was far from universal in 1905, and the success of Einstein's explanation of the well-documented existence of Brownian motion did much to convince a number of distinguished scientists that such things as atoms actually exist. 
%
%\item  Incidentally, 1905 is the same year in which Einstein set forth his theory of relativity and his quantum explanation for the photoelectric effect. 
%
%\item  Any single one of his 1905 contributions would have brought him recognition by his fellow physicists. 
%
%\item  Today, a search in a university library under the subject heading `Brownian motion' is likely to turn up dozens of books on the stochastic process called Brownian motion and few, if any, on the irregular movements observed by Robert Brown. 
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%\item  The literature on the model has far surpassed and overwhelmed the literature on the phenomenon itself!
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%\end{itemize}
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\begin{frame}{8.3. The Brownian Motion Stochastic Process }

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\begin{itemize}

\item  In terms of our general framework of stochastic processes, the Brownian motion process is an example of a {\color{red}continuous-time, continuous-state-space} Markov process. 

\item  Let $B(t)$ be the $y$ component (as a function of time) of a particle in Brownian motion. 

\item  Let $x$ be the position of the particle at time $t_0$; i.e., $B(t_0) = x$. 

\item  Let $p(y,t\mid x)$ be the probability density function, in $y$, of $B(t_0 + t)$, given that $B(t_0) = x$. 

\item  We postulate that the probability law governing the transitions is stationary in time, and therefore $p(y,t\mid x)$ does not depend on the initial time $t_0$.

\end{itemize}

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\begin{itemize}

\item  Since $p(y,t\mid x)$ is a probability density function in $y$, we have the properties\begin{eqnarray} p(y, t\mid x) \ge 0 \text{ and } \int_{-\infty}^{\infty} p(y, t\mid x) dy =1.
\label{eq8-1}
\end{eqnarray}

\item  Further, we stipulate that $B(t_0 + t)$ is likely to be near $B(t_0) = x$ for small values of $t$.

\item  This is done formally by requiring that
\begin{eqnarray} 
\lim\limits_{t\to 0} p(y,t\mid x) = 0 \text{ for } y\neq x.\label{eq8-2}
\end{eqnarray}

\item  From physical principles, Einstein showed that $p(y,t\mid x)$ must satisfy the partial differential equation
\begin{eqnarray} 
\frac{\partial p}{\partial t} = \frac{1}{2} \sigma^2 \frac{\partial^2 p}{\partial x^2}.
\label{eq8-3}
\end{eqnarray}
\end{itemize}

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\begin{frame}{8.5. The diffusion equation }

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\begin{itemize}
\item  This is called the {\color{red}diffusion equation}, and $\sigma^2$ is the diffusion coefficient, 
which Einstein showed to be given by $\sigma^2 = RT/Nf$, where $R$ is the gas constant, $T$ is the temperature, $N$ is Avogadro's number, and $f$ is a coefficient of friction. 


\item  By choosing the proper scale, we may take $\sigma^2 = 1$. 

With this choice, we can verify directly that\begin{eqnarray} p(y,t\mid x) = \frac{1}{\sqrt{2\pi t}} \exp\left( -\frac{1}{2t}(y-x)^2\right).
\label{eq8-4}
\end{eqnarray}
is a solution of (\ref{eq8-3}). 

\item  In fact, it is the only solution under the conditions (\ref{eq8-1}) and (\ref{eq8-2}). 

\end{itemize}

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\begin{itemize}

\item  We recognize (\ref{eq8-4}) as a normal probability density function whose mean is $x$ and whose variance is $t$. 
That is, the position of the particle $t$ time units after observations begin is normally distributed. 

\item  The mean position is the initial location $x$, and the variance is the time of observation $t$.

\item  Because the normal distribution will appear over and over in this chapter, we are amply justified in standardizing some notation to deal with it. Let
\begin{eqnarray} 
\phi(z) = \frac{2}{2\pi} e^{-z^2/2}, -\infty < z < \infty,\label{eq8-5}
\end{eqnarray}be the standard normal probability density function, 



\end{itemize}

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\begin{itemize}

\item  Let $\Phi(z)$ be the corresponding {\color{red}cumulative distribution function},
\begin{eqnarray} 
\Phi(z) = \int_{-\infty}^{z} \phi(x)dx, 
\label{eq8-6}
\end{eqnarray}
\item A small table of the cumulative normal distribution appears here. 

\begin{table}[ht!]
\centering
\begin{tabular}{cc|cc}
$x$ & $\Phi(x)$ & $x$ & $\Phi(x)$ \\ \hline
$-3$ & 0.00135 & $-2.236$ & 0.01 \\ 
$-2$ & 0.02275 & $-1.96$ & 0.025 \\ 
$-1$ & 0.1587 & $-1.645$ & 0.05 \\ 
$0$ & 0.5000 & $-1.282$ & 0.10 \\ 
$1$ & 0.8413 & $1.282$ & 0.90 \\ 
$2$ & 0.97725 & $1.645$ & 0.95 \\ 
$3$ & 0.99865 & $1.96$ & 0.975 \\ 
\end{tabular}
\end{table}

\end{itemize}

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\begin{itemize}

\item  Let $\phi_t(z)$ and $\Phi_t(z)$ be the probability density function and cumulative distribution function, respectively,
\begin{eqnarray} 
\phi_t(z) &=& \frac{1}{t} \phi(z/\sqrt{t}), \label{eq8-7} \\
\Phi_t(z) &=& \int_{-\infty}^{z} \phi_t(x)dx = \Phi(z/\sqrt{t}), \label{eq8-8}
\end{eqnarray}for the normal distribution with mean zero and variance $t$. 

\item  In this notation, the transition density in (\ref{eq8-4}) is given by
\begin{eqnarray} 
p(y,t\mid x)=\phi_t(y-x). 
\label{eq8-9}
\end{eqnarray}



\end{itemize}

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%\begin{itemize}
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%\item  We calculate the probability 
%\begin{eqnarray*} 
%\mathbb{P}\{ B(t) \le y\mid B(0)=x\} =  \Phi\left( \frac{y-x}{\sqrt{t}} \right) .
%%\label{eq8-9}
%\end{eqnarray*}
%
%\item  The transition probability density function in (\ref{eq8-4}) or (\ref{eq8-9}) gives only the probability distribution of $B(t)-B(0)$. 
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%\item  The complete description of the Brownian motion process with diffusion coefficient $\sigma^2$ is given by the following definition.
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\begin{itemize}

\item  {\color{red}Definition}. Brownian motion with diffusion coefficient $\sigma^2$ is a stochastic process $\{B(t); t\ge 0\}$ with the properties:

\begin{enumerate}\item  Every increment $B(s+t)-B(s)$ is normally distributed with mean zero and variance $\sigma^2t$; $\sigma^2 > 0$ is a fixed parameter.\item  For every pair of disjoint time intervals $(t_1,t_2], (t_3,t_4]$, with $0\le t_1 < t_2 \le t_3 < t_4$, 
the increments $B(t_4) - B(t_3)$ and $B(t_2) - B(t_1)$ are independent random variables, 
and similarly for $n$ disjoint time intervals, where $n$ is an arbitrary positive integer.\item  $B(0) = 0$, and $B(t)$ is continuous as a function of $t$.
\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  The definition says that a displacement $B(s + t) - B(s)$ is independent of the past, or alternatively, if we know $B(s) = x$, then no further knowledge of the values of $B(\tau)$ for past times $\tau < s$ has any effect on our knowledge of the probability law governing the future movement $B(s + t) - B(s)$. 

\item  This is a statement of the Markov character of the process. 

\item  We emphasize, however, that the independent increment assumption (b) is actually more restrictive than the Markov property. 

%\item  A typical Brownian motion path is illustrated in the Figure.

%\item  The choice $B(0) = 0$ is arbitrary, and we often consider Brownian motion starting at $x$, for which $B(0) = x$ for some fixed point $x$. 

\item  For Brownian motion starting at $x$, the variance of $B(t)$ is $\sigma^2t$, and $\sigma^2$ is termed the variance parameter in the stochastic process literature. 

%\item  The process $\tilde{B}(t) = B(t)/\sigma$ is a Brownian motion process whose variance parameter is one, the so-called {\color{red}standard Brownian motion}. 
%
%\item  By this device, we may always reduce an arbitrary Brownian motion to a standard Brownian motion; for the most part, we derive results only for the latter.  


\end{itemize}

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\begin{itemize}

\item  Using the continuity of the trajectories of Brownian motion and the symmetry of the normal distribution, we will determine a variety of interesting probability expressions for the Brownian motion process. The starting point is the reflection principle.

\item  Let $B(t)$ be a standard Brownian motion. Fix a value $x > 0$ and a time $t > 0$. 
Bearing in mind the continuity of the Brownian motion, property (c) of the definition, consider the collection of sample paths $B(u)$ for $u \ge 0$ with $B(0) = 0$ and for which $B(t) > x$. 

\item  Since $B(u)$ is continuous and $B(0) = 0$, there exists a time $\tau$, itself a random variable depending on the particular sample trajectory, at which the Brownian motion $B(u)$ first attains the value $x$.


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\begin{itemize}

\item  We next describe a new path $B^*(u)$ obtained from $B(u)$ by reflection. 

\begin{figure}
\centering
\includegraphics[height=0.7\textheight, width=0.9\textwidth]{pic/reflection-principle.png}
% \caption{ }
\end{figure}

\end{itemize}

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\begin{itemize}

\item  For $u > \tau$, we reflect $B(u)$ about the horizontal line at height $x > 0$ to obtain $B^*(u)$. Note that $B^*(t) < x$ because $B(t) > x$.
\begin{eqnarray*}B^*(u) = \left\{\begin{array}{ll}
B(u), & \text{ for } u \le \tau, \\
x - [B(u)- x], & \text{ for }u > \tau.
\end{array}\right.
\end{eqnarray*}


\item  Because the conditional probability law of the path for $u > \tau$, given that $B(\tau) = x$, is symmetric with respect to the values $y > x$ and $y < x$, and independent of the history prior to $\tau$, the reflection argument displays for every sample path with $B(t) > x$ two equally likely sample paths, $B(u)$ and $B^*(u)$, for which both
\begin{eqnarray*}
\underset{0\le u\le t}{\max} B(u)>x \text{ and } \underset{0\le u\le t}{\max} B^*(u)>x. 
\end{eqnarray*}

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\begin{itemize}


\item  Conversely, by the nature of this correspondence, every sample path $B(u)$ for which $\underset{0\le u\le t}{\max} B(u)>x$ results from either of two equally likely sample paths, exactly one of which is such that $B(t) > x$. 

\item  The two-to-one correspondence fails only if $B(t) = x$, but because $B(t)$ is a continuous random variable (normal distribution), 
we have $\mathbb{P}\{B(t) = x\} = 0$, and this case can be safely ignored. 

\item  Thus, we conclude that%\begin{eqnarray*}$\mathbb{P}\left\{ \underset{0\le u\le t}{\max} B(u) > x \right\} = 2\mathbb{P}\{ B(t) > x\}. 
$%\end{eqnarray*}
\item  In terms of the maximum process defined by 
%\begin{eqnarray}
$M(t) = \underset{0\le u\le t}{\max} B(u)$, 
%\label{eq8-19}
%\end{eqnarray}
and using the notation set forth in (\ref{eq8-8}), we have%\begin{eqnarray}
$\mathbb{P}\{ M(t) > x\} = 2[1 -\Phi_t(x)]$.
%\label{eq8-20}
%\end{eqnarray}


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